2021 WSMO Accuracy Round Problems/Problem 4

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Problem 4

A 12-hour clock has a minute hand that is the same length as the second hand, and an hour hand half the length of the minute hand. In a day, the tip of the minute hand travels a distance of $m,$ the tip of the second hand travels a distance of $s,$ and the tip of the hour hand travels a distance of $h.$ The value of $\frac{m^2}{hs}$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a+b$.

Solution 1

WLOG, assume that the length of the minute hand is 2. This means that the length of the second hand is 2 and the length of the hour hand is 1. In a day, there are 24 hours, which means that the minute hand travels $24\cdot4\pi$ each day. Also, since there are $24=2\cdot12$ hours, the hour hand travels $2\cdot2\pi=4\pi$ each day. Finally, there are $24\cdot60$ minutes in a day, which means that the second hand travels $24\cdot60\cdot4\pi$ each day. Thus, the final answer is \[\frac{(24\cdot4\pi)\cdot(24\cdot4\pi)}{(4\pi)\cdot(24\cdot60\cdot4\pi)}=\frac{(\cancel{24}\cdot\cancel{4\pi})\cdot(24\cdot\cancel{4\pi})}{(\cancel{4\pi})\cdot(\cancel{24}\cdot60\cdot\cancel{4\pi})}=\frac{24}{60}=\frac{2}{5}\Longrightarrow 2+5=\boxed{7}.\] ~pinkpig

Solution 2

Let the distance traveled by one revolution of the minute hand tip be $C$. Note that $m=24C$, $s = 60 \cdot 24 C$, and $h=2(\frac{C}{2}) = C$. Our desired expression becomes: \[\frac{(24C)^2}{(C)(60 \cdot 24 C)}\] \[=\frac{24^2}{60 \cdot 24}\] \[=\frac{2}{5}\]

This gives us an answer of $2 + 5 = \boxed{7}$.

~BigKahuna227